Answer:
The concentration in mol/L = 4.342 mol/L
Explanation:
Given that :
mass of sodium chloride = 25.4 grams
Volume of the volumetric flask = 100 mL
We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol
and number of moles = mass/molar mass
The number of moles of sodium chloride = 25.4 g/58.5 g/mol
The number of moles of sodium chloride = 0.434188 mol
The concentration in mol/L = number of mol/ volume of the solution
The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L
The concentration in mol/L = 4.34188 mol/L
The concentration in mol/L = 4.342 mol/L
[Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.
<h3>What is telleurium?</h3>
Tellurium is a noble gas element that is non-reactive in nature due to complete outermost shell. It has atomic number of 52 which means that it has 52 number of electrons.
So we can conclude that [Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.
Learn more about noble gas here: brainly.com/question/13715159
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Answer: An atom in an excited state contains more of kinetic energy than the same atom in the ground state.
Explanation:
Kinetic energy is the energy acquired by an object due to its motion. And, thermal energy is the internal energy of an object arisen because of the kinetic energy present within the molecules of the object.
Potential energy is the energy acquired by an object due to its position.
The total energy present at the center of mass of an object is known as mass-energy.
So, when an atom gets excited then it means it is gaining kinetic energy due to which it moves from its initial position after getting excited.
Thus, we can conclude that an atom in an excited state contains more of kinetic energy than the same atom in the ground state.
In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
